सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

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(Sūtra × H cos δ) / R = Kalā = (H cos h . H cos δ) / R = ∴
(H cos h . √(R² — x²)) / R
But Kalā is the Koti of the fifth latitudinal triangle of which
H sin δ is Bhuja. Hence (Kalā × H sin ϕ) / (H cos ϕ) = H sin δ = x
ie. (H cos h √(R² — x²)) / R × (H sin ϕ) / (H cos ϕ) = x; but (H sin ϕ) / (H cos ϕ) = s / 12
∴ Squaring both sides [(R² — H sin² h) (R² — x²)] / R² × s² / 12² = x²
∴ 12² R² x² = s² R² (R² — H sin² h) — s² x² (R² — H sin² h)
ie. x² {(12² R² + s² (R² — H sin² h)} = s² R² (R² — H sin² h)
∴ x² = [s² R² (R² — H sin² h)] / [12² R² + s² (R² — H sin² h)]
= R² / [(12² R²) / (s² (R² — H sin² h)) + 1] ∴ x = R / √[(12² R²) / (s² (R² — H sin² h)) + 1] =
H sin δ as given. From H sin δ, the method of obtaining
λ is clear from the formula (H sin λ H sin ω) / R = H sin δ. In
the given numerical example h = 5 nādīs = 360° / 12 = 30°
since 60 nadis of time correspond to 360°.
Thus H sin h = R / 2 ; the remaining work follows.
Verse 86. Another question.
When the Sun is on the prime-vertical the gnomonic
shadow is noted to be 16 inches. The Unnatakāla is 8
nādis. If you could give the H sin δ and s, I shall con-
sider you nothing short of one who is an adept in solving
the totality of the diurnal problems.
Verses 87 and 88. Answer to the question.